help-26

[keen matrix]

for example, 2/4 and 1/2 are the same fraction

WLOG, they are saying, well if sqrt(2) is rational, let's at least put it in lowest terms

from there, the contradiction arises

[drifting swift]

are you trying to ask for the motivation of requiring them to be coprime

as in "why do we bother imposing this requirement"

if so: every fraction has a unique lowest-terms form. this is usually what is most convenient to work with -- and the idea of the proof as a whole is to run ourselves into a contradiction with that

(by, essentially, finding an even lower-terms form than the one in lowest terms -- mind you what i just said is informal)

[dusky ridge]

nvm got it

thank you

.close

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[topaz sinew]

@fair furnace Has your question been resolved?

[sharp pagoda]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[fair furnace]

We know there is C(P, 2) pairs of players

that is P(P-1)/2

[sharp pagoda]

Yea

[fair furnace]

At each table is C(C, 2) pairs

[sharp pagoda]

Yea but let him try to do it

And i can handle this

[fair furnace]

I can do 2P(P-1)/(T*2C(C-1))

P(P-1)/TC(C-1)

The problem is that, we assume that all pairs are new

but this isn't the case

so P(P-1)/TC(C-1) = 1024(1023)/(24*6(5)) = 1454.9333

So 1455, is just lower bound, not the solution

[sharp pagoda]

Yes

[fair furnace]

Is there way to get the exact solution?

[sharp pagoda]

I have an idea but it might be wrong

We check that in 1024 people, there is C(1024,6) when sectos of people

Which is 1,57795e15

[fair furnace]

ok

[sharp pagoda]

Hmm

[fair furnace]

the problem is that I am not even able to write a greedy backtracking solution

[sharp pagoda]

I think this is not the solution

These numbers must have some connection

Hmm i cant figure it out

Maybe @wintry meteor

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@fair furnace Has your question been resolved?

[mint crescent]

$1024 \geq 24 \cdot 6$

[thorny flame]

Civil Service Pigeon

[fair furnace]

yeah that means not all players can be seated in a single round

[topaz sinew]

@fair furnace Has your question been resolved?

[sharp pagoda]

this is hard

i think you need repetitive combination

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@fair furnace Has your question been resolved?

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[vestal cape]

how do i do the d's for section 1.7

[sweet shard]

if f(x) = sqrt(x), then -sqrt(x) + 4 = -[sqrt(x)] + 4 = ?

[vernal vale]

it may help to start with a list of parent functions

if you can locate that in the book

[vestal cape]

I have all the parent functions figured out

[vernal vale]

but riemann is describing basically what you must do

[vestal cape]

i lowkey dont understand what riemann just said

[vernal vale]

well how do you feel about part b

describe a sequence of transformations

[vestal cape]

figured a-c out

just dk how to do d

[vernal vale]

for part d, you just need to put part b into math language

meaning, some equation, in terms of f

if you have the parent function $f(x)=x^2$, then $h(x) = x^2-9 = f(x) - 9$, maybe

[thorny flame]

jan Niku

[vernal vale]

i imagine your part b looked something like, we start with $x^2$, and move it vertically down 9 units?

[thorny flame]

jan Niku

[vestal cape]

yes

[vernal vale]

so you just say this in math terms

using the parent function f = x^2

does that help?

[vestal cape]

isnt it already written like that but with h(x)

[vernal vale]

it's an easy problem, good to use as an example

maybe something like $h(x) = -(x+2)^2 + 3$ is more interesting

[thorny flame]

jan Niku

[vernal vale]

whats the parent function here?

[vestal cape]

f(x) = x^2

[vernal vale]

whats your part b for problem 119

yea

what were your transformations

[vestal cape]

reflection over x-axis, left 2, up 3

[vernal vale]

its pretty easy to say this in math language

$h(x) = -(x+2)^2 + 3 = - f(x+2) + 3$

[thorny flame]

jan Niku

[vernal vale]

where $f(x) = x^2$ is the parent function

[thorny flame]

jan Niku

[vestal cape]

where did the square go

[vernal vale]

f is handling that

f takes the input and squares it

so $(x+2)^2 = f(x+2)$

[thorny flame]

jan Niku

[vestal cape]

ohhh

gotcha

[vernal vale]

does this make sense

[vestal cape]

yeah

[vernal vale]

you can try tackling some of the others

ping if you get into hassles

[vestal cape]

so basically you just drop whatever is with the x in the parent function

and put f

[vernal vale]

yea i mean in math language

any transformation to the independent variable, that goes as part of the input to f, the parent function

[vestal cape]

so 2|x-5|+3 would be 2f(x-5)+3?

[vernal vale]

here, thats a left translation of 2

[vernal vale]

where f(x) the parent function is |x|, yea

[vestal cape]

gotcha

[vernal vale]

youll have to identify it, as part of this problem, i believe

[vestal cape]

alr

thanks

[vernal vale]

[vestal cape]

i need help with 76

[vestal cape]

i got .4(50-x)+x for a

[vernal vale]

whats a?

[vestal cape]

teacher talked about that in class

.4(50-x)+x

[vernal vale]

i mean, how do you define a

[vestal cape]

like f(x)?

[vernal vale]

i mean, you say you have some equation for a, im asking what a is

just because im a stickler about units

since theyre so helpful in problems like this

[vestal cape]

question 761?

76a?

[vernal vale]

yea

[vestal cape]

f(x) = .4(50-x)+x?

am i trippin

[vernal vale]

hmm maybe i will work it out and compare

[vestal cape]

thats what the math teacher said 76a was

[vernal vale]

hmm i get something different

[vestal cape]

hmm

[vernal vale]

but maybe it ends up equivalent

i figure

you have 50*(0.4) to begin with

then you remove X liters

so you have 50*(0.4) - X(0.4)

[fiery canopy]

You're finding the mixture after a amout of x liters is removed

[vernal vale]

then, you add X liters of 100%

so 50*0.4 - X*(0.4) + X*(1.0)

[fiery canopy]

The container has 50 liters of a 40% acid concentration, so the initial amount of acid is 0.40×50=20L

Does that makes sense or I'm just hella confused rn

[vernal vale]

it makes sense to factor the X

[vestal cape]

uh

[vernal vale]

we get 50 * 0.4 + X*[ 1.0 - 0.4 ]

[vestal cape]

well if you add 0% acid

you get .4(50-0)+0

so just 20

thats correct

[fiery canopy]

Yaa

[vestal cape]

if you add 10 liters you get 26

so i think the .4(50-x)+x = f(x) is correct

[vestal cape]

oh

thats just for the initial amount of acid

but we need an equation for any amount of it added

[fiery canopy]

Aren't they suppose to simply?

So it's gonna be like (20-0.4x) + x = f(x)

[vestal cape]

ig any would work

[fiery canopy]

It seems correct

[vestal cape]

yeah

[fiery canopy]

Yaa xD

[vestal cape]

they are both the same

[fiery canopy]

Ahh makes sense lol

[vernal vale]

[vestal cape]

so uh

moving on to 76 b

how are you supposed to find that

[vernal vale]

do you know what domain and range mean?

[vestal cape]

yeah

[vernal vale]

I mean, in the context of this problem

[vestal cape]

no

[vernal vale]

Well domain is the independent variable right

here thats x

[fiery canopy]

Yeah

[vestal cape]

oh yeah

[fiery canopy]

Yep

[vernal vale]

are there restrictions on x

[vestal cape]

ye

[vernal vale]

what are they

[fiery canopy]

Domain is x and range is y

[vernal vale]

are there values that dont make sense for x

[vestal cape]

you can only have 0-30 liters for x

[fiery canopy]

f(x) is y

[vernal vale]

i think maybe 50

[fiery canopy]

And the equation is the domain

[vernal vale]

the original volume of the barrel

like

it doesnt make sense to remove more than we started with

maybe thats what you meant

[vestal cape]

if 40% is already acid doesnt that only leave 30l

[fiery canopy]

Wait- what the hell is x ToT

[vernal vale]

yea, but

this is where units are really important

x is in terms of volume of liquid

regardless of the concentration

X is Liters

we might also say

[fiery canopy]

I think the domain is [0,50]

[vernal vale]

it is

but did you want to leave that for pidek

[fiery canopy]

Because the container can't hold more or less than the container

[vestal cape]

oh wait

does the acid count as part of the liters you can remove

like you can remove acid then put more in

[vernal vale]

so X is some amount of liters, regardless of acid

totally remove acid or concentration from your mind

X is just the amount of volume we remove from the barrel, and replace it with some other liquid

[vestal cape]

ah

got it

[vernal vale]

So, because the barrel only holds 50 Liters

what kind of restrictions make sense?

does it make sense to remove -500 liters?

[vestal cape]

nope

[vernal vale]

does it make sense to remove 500 Liters from a 50 liter barrel?

So what kind of restrictions should be placed on X

[vestal cape]

0-50

[vernal vale]

yea

these are domain restrictions

you know its domain because domain means independent variable

[vestal cape]

yeah

[vernal vale]

the x in f(x)

now you have to figure out the range restrictions

these come from the domain restrictions, usually

normally, youll check each end.

you check the lowest of the domain, and the highest of the domain

[vestal cape]

would it be the same as the domain?

[vernal vale]

not usually no

here it definitely won't

because your independent variable is X in terms of Liters

but your output a(x) (or whatever you want to call it)

[vestal cape]

is y here the amount of liters in the final mixutre?

mixture

[vernal vale]

is in terms of concentration

its the amount of liters per total barrel volume

so amount of pure acid in liters, divided by the volume of the barrel

the specifics aren't that important

you should know that usually, to get range, you just plug in the extremes of your domain

you will check a(0), and a(50)

and that will tell you how wide your range can be

the minimum, and maximum, concentration

when we replace nothing a(0), and when we replace everything a(50)

[vestal cape]

so [20,50]?

[vernal vale]

this is right if the units of a are liters

yea, that seems right

[vestal cape]

awesome

76c now

[vernal vale]

What do you think?

so....you have a function a(x)

[vernal vale]

this seems to say

a(x) gives you liters

but now were coming it from the other direction

we know a(x), kind of

but we dont know x

say we're still talking about the same 50 liter barrel

[vestal cape]

would it just be y=25

[vernal vale]

yea

so a(x) = 25

can you solve it for x?

[vestal cape]

hm

dont think so

wait

8 1/3?

[vernal vale]

,calc 50 * 0.4 + 25/3 * 0.6

[thorny flame]

Result:

25

[vernal vale]

[vestal cape]

oh shoot

i thought i was dead wrong

lmao

[vernal vale]

nah i think you got it

[vestal cape]

awesome

[vernal vale]

i gotta stop helping with math

can i suggest, if you have other problems

open a new channel

so it puts you at the top of the list

[vestal cape]

alr

thanks

[vernal vale]

[vestal cape]

.close

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[tranquil gust]

i dont even know where to start

[stiff knoll]

compare x when $p = 10$ to when $p = 11.80$

[thorny flame]

Médicis

[tranquil gust]

x goes from 12 to ~14.915

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@tranquil gust Has your question been resolved?

[tranquil gust]

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[shy raft]

laushan says hi

laushaaaan

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[sonic vine]

I eliminated oppption c and d but how do I know if its a or b

[slim mountain]

A isnt product rule

its chain rule

yk its b bc thasts the only one with a product

@sonic vine

[sonic vine]

uh

[slim mountain]

uh

[exotic nest]

uh

[sonic vine]

i am a little confused

[slim mountain]

its not

[sonic vine]

how do i know if its chain rule

oh my god

[slim mountain]

arcsin (x) cos (x)

[sonic vine]

om stupid

lol

[slim mountain]

ur good

[sonic vine]

i been looking at it wrong

lol ty guys

[slim mountain]

yeah its written weird

i dont think we helped much lol

[sonic vine]

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[open bridge]

Hey! I'm supposed to explain what I did but the thing is I don't know how what I did works. Could someone please explain to me how constructing these circles proves that the new line formed is parallel to segment QT?

[pseudo bear]

What steps did you take?

[open bridge]

I first used Point Q to Point T has the radii of a circle with T has the center point

then extended segment QT so it intersects the circle

from where they intersected I used that as the new center point and line QT as an radii to form a new circle

then I did the same process on the other side using Q

And then I used a straightedge to make a line where the two circles met on both sides

finally I used a straightedge to connect the to lines formed making sure it goes through R

And then the arch's in the middle I dont think help me in any way but for that I used Q as an center point and QR as the radii and did the same thing with T

[pseudo bear]

For the first part, is that the semicircle you drew?

[open bridge]

Yeah

[pseudo bear]

OK, for the second part, it doesn't seem like QT is extended to intersect the semicircle on the right side.

[open bridge]

oh the line dosent show up in the photo my pencil was running out so the lines faint

[pseudo bear]

Oh, OK.

For step five, which circles are the two circles that meet?

You have four circles at that point in the process.

[open bridge]

Does this help? the ones in the blue is where they met and I made lines

[pseudo bear]

What I'd recommend is to set the compass to radius QR, center it on Q, and draw the circle. Then set the compass to radius TR, center it on T, and draw the circle. Then, you can draw a line between the two circle intersections split QT below R at a right angle.

It's kind of like a bisector, only that R isn't guaranteed to be halfway between Q and T.

[open bridge]

Okay thanks!

[pseudo bear]

Then we need to make a right angle to the line R'R where R' is the point under R on QT.

Let's see.

I'm not sure if this is constructible, but perhaps set the compass to length R'R and center it on T to draw a circle. Then set the compass to length R'T and center it on R to draw a circle.

The circles will intersect where you have your upper right corner of the rectangle.

Then, you can do the same with T replaced with Q to get the upper left corner of the rectangle.

Then, you can draw a line between the top corners.

[open bridge]

So something like this?

[pseudo bear]

Oh, wait.

Apparently, you can't set the compass to a certain radius and then lift it up.

[open bridge]

oh

Welp Ima sleep on this thanks for the help!

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[pseudo bear]

No problem.

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[terse abyss]

$\log_a{a}=1\forall a \in R$ or $R_+$ or am i totally off?

[thorny flame]

Pixelius

[clear python]

Doesn't work for a = 1

To begin with

Also I don't think log is defined for negative bases

So $\mathbb{R}^+ \setminus {1}$ ig

[thorny flame]

Xavier 🌺

[terse abyss]

Got it thanks

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[faint ore]

I don't get the underlined bits. Why is it true?

[faint ore]

Guys pls ping me to replyyy

Help is appreciated

ping to reply

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@faint ore Has your question been resolved?

[faint ore]

<@&286206848099549185>

[somber oasis]

i have arrived

matcha & caramel

its been a while

has ur handwriting gotten any better

but anyways

i guess u want me to ping u

@faint ore

bcause each phi step removes at most one factor of 2 and the weight function w(n) exactly counts how many 2s appear in the phi chain, we get k >= w(n) and since by induction w(n) >= log base 3 of n it follows that n <= 3^k

[topaz sinew]

@faint ore Has your question been resolved?

[faint ore]

But how does w(n) count the number of 2s

[faint ore]

Improved drastically these last few days 😄

Sorry I was eating just now brotha

Just DM me

[faint ore]

Cause I can't rlly talk rn

[somber oasis]

because w is defined by w(ab)=w(a)+w(b), w(2)=1, and w(p)=w(p-1) when u iteratively replace every odd prime p in n by (p-1) just like phi does ueventually get only 2s and the add up w(n) exactly equals how many factors of 2 appear in that whole process

wrong reply

but u get the point

[faint ore]

Hmm

It's kinda confusing

Yk what can we continue this like 4 hours later I have an exam in 1 hour

Thank u tho

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[burnt swift]

4. Let $$F(x,y,z) = x^3 + y + e^{xy(z-1)} - \cos(xyz) + z$$

   a) Show that $F(x,y,z) = 0$ defines an implicit function $z = \varphi(x,y)$ in a neighborhood of $(0,2,-2)$. \
   b) Let $h(x,y) = \varphi(x+y, x - y).$ Find $\nabla h (1,-1)$

[thorny flame]

Renato

[burnt swift]

need help with b

[prisma mesa]

define u = x+y and v = x-y and apply chain rule

[burnt swift]

so

h(s,t) = phi(s + t, s - t)

now what?

[prisma mesa]

what

where did u get s and t from

[burnt swift]

I prefer to use s instead of u

any problem?

[prisma mesa]

well, what you wrote isnt entirely correct, but nvm, well fix that

what's grad h(1, -1)?

write it explicitly as a vector

[burnt swift]

◇ h(1,-1) = (hs(1,-1) , ht(1,-1))

[prisma mesa]

I'd use x and y ◇ h(1,-1) = (h_x(1,-1) , h_y(1,-1))

but okay, now we know that we have to find the partial derivative of h wrt x

[burnt swift]

you said use s and t

well he said use u and v which is the same regardless, the gradient would be in terms or u and v

that's is what I am saying

now he is using the gradient being in terms of x and y, what is going on . . .

[prisma mesa]

tbh i dont see what u did here at all

if s = x + y and t = x - y, how did u get that?

or why

[burnt swift]

I dont understand

h(x,y) = phi(x+y, x - y)

you use s = x + y and t = x - y, then what do you use here? h(?,?) = (s, t)

h(s,t) = (s,t) is just the identity function

[prisma mesa]

h(x, y) = phi(s, t) is good enough
Now what the gradient literally is is the vector of partial derivatives wrt x and wrt y. So we do that on both sides. The point is that you apply chain rule

you can kind of imagine it like this h(x, y) = phi(s(x, y), t(x, y))

s and t are dependent on x and y

the gradient is the vector of partial derviatives wrt x and wrt y, we cant change that (or at least not so simply)

so now we need to compute exactly that

[burnt swift]

the partials of h

[prisma mesa]

$h_x(x,y) = \frac{\partial}{\partial x} \varphi(s(x,y), t(x,y))$

[thorny flame]

MathIsAlwaysRight

[prisma mesa]

now we just apply chain rule

[burnt swift]

so $\pdv{h}{x} = \pdv{\varphi}{s} \cdot \pdv{s}{x} + \pdv{\varphi}{t} \cdot \pdv{t}{x}$ and $\pdv{h}{y} = \pdv{\varphi}{s} \cdot \pdv{s}{y} + \pdv{\varphi}{t} \cdot \pdv{t}{y}$

[thorny flame]

Renato

[prisma mesa]

perfect

that's exactly what i was trying to communicate, sorry for the miscommunication

okay now ds / dx, dt / dx, ds / dy and dt / dy simplify very nicely

and whats left is computing d phi / ds and d phi / dt

[burnt swift]

(s(x,y) ,t(x,y)) = (x+y, x - y)

∂s/∂x =1

∂s/∂y = 1

∂t/∂x = 1

∂t/∂y = -1

[prisma mesa]

yep, thats it

now note that phi(s, t) is just like the phi(x, y) in z = phi(x, y), because those x and y dont depend on s and t (they were used in defining phi, they are not the same as the ones used in part b)

[burnt swift]

so $\pdv{h}{x} = \pdv{\varphi}{s} \cdot \pdv{s}{x} + \pdv{\varphi}{t} \cdot \pdv{t}{x}$ and $\pdv{h}{y} = \pdv{\varphi}{s} \cdot \pdv{s}{y} + \pdv{\varphi}{t} \cdot \pdv{t}{y}$ \ $\pdv{h}{x} = \pdv{\varphi}{s} \cdot 1 + \pdv{\varphi}{t} \cdot 1$ and $\pdv{h}{y} = \pdv{\varphi}{s} \cdot 1 + \pdv{\varphi}{t} \cdot (-1)$

[thorny flame]

Renato

[prisma mesa]

so computing dphi / ds is gonna be just like computing dphi / dx in the original defn

(and btw by d i mean that \partial symbol when typing in text)

[burnt swift]

what is your point

[prisma mesa]

my point is that you can now focus purely on this part and find phi_x and phi_y and those willl be the phi_s and phi_t

[prisma mesa]

yeah, and u can use exactly that

[burnt swift]

F(0,2 phi(0,2)) = F(0,2,-2)

phi(0,2) => x + y = 0 and x - y = 2

2x = 2 => x = 1

y = -1

[prisma mesa]

note that we are going for the gradient at x = 1 and y = -1, which is s = 0 and t = 2

[prisma mesa]

which is just like the x = 0 and y = 2 in part a)

[prisma mesa]

and so u can really just directly compute this at x = 0 and y = 2

[burnt swift]

we need the gradient of F

[prisma mesa]

if I wanted to, i could really rephrase all of a) without changing it like this:

Let F(s, t, z) = s^3 + t + e^(st(z-1)) - cos(stz) + z
Show that F(s, t, z) = 0 defines an implicit function z = phi(s, t) in a neighborhood of (0, 2, -2)

[prisma mesa]

yeah, and you can compute partial derivatives of F, cant you?

you need the partials wrt x and wrt y really

it doesnt matter what variables you use in the definition

whether they are s, t or x, y

the x and y in a) are different from the x and y in b)

[burnt swift]

also wrt z

[prisma mesa]

yeah, that too

[burnt swift]

so we need all of the partials, the gradient

[prisma mesa]

yep

and thats all you need

[burnt swift]

is just that phi(s,t) = phi(x,y)

[prisma mesa]

they are not really equal, but you can treat them the same if there is no relation between s, t and x, y

in b) we had that relation. But the x and y in b are different from the x and y in a)

[burnt swift]

,, F_x = 3x^2 + (y(z-1))e^{xy(z-1)} + yz\sin(xyz)

[thorny flame]

Renato

[prisma mesa]

seems good

[burnt swift]

you sure?

,, F_y = 1 + (x(z-1))e^{xy(z-1)} + (xz)\sin(xyz)

[thorny flame]

Renato

[prisma mesa]

about what?

about why we can treat phi(s, t) just like phi(x, y) in some sense?

[burnt swift]

the partials

i mean

[prisma mesa]

yea, i think u differentiated it correctly if u mean that

[burnt swift]

,, F_z = (xy)e^{xy(z-1)} + (xy)\sin(xyz) + 1

,w partial derivatives x^3+y+e^(xy(z-1)) - cos(xyz) + z at (0,2,-2)

[prisma mesa]

yep, seems alright

[burnt swift]

i messed up the last one

[prisma mesa]

oh, you forgot to multiply e^xy(z-1) with xy

[thorny flame]

Renato

[burnt swift]

ye, too much mental math

my brain collapsed

[burnt swift]

yeah

there is no relation of s and t with the z = phi(x,y)

[prisma mesa]

okay, now we can evaluate them at (0, 2, -2)

[prisma mesa]

exactly

thank god x is 0, otherwise it wouldve been really messy

it's not that hard to plug in by hand btw, almost everything disappears

due to x = 0

in Fy and Fz, only the 1s survive

in Fx, only y(z-1) survives and thats just 2(-2-1) = -6

[burnt swift]

,, F_x = 3x^2 + (y(z-1))e^{xy(z-1)} + yz\sin(xyz) \ F_y = 1 + (x(z-1))e^{xy(z-1)} + (xz)\sin(xyz) \ F_z = (xy)e^{xy(z-1)} + (xy)\sin(xyz) + 1

[thorny flame]

Renato

[burnt swift]

Fx(0,2,-2) = 0 + (2*-3)e^(0) -4sin(0)
Fx(0,2,-2) = 0 + -6 + 0 = -6

Fy(0,2,-2) = 1 + 0 + 0 = 1

Fz(0,2,-2) = 0 + 0 + 1

Fz(0,2,-2) = 0 + 0 + 1 = 1

,w partial derivatives of (x^3+y+e^(xy(z-1)) -cos(xyz) + z) at (0,2,-2)

[prisma mesa]

seems good

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by renatoxbz

[burnt swift]

,, \varphi_x (0,2) = -\frac{-6}{1} = 6 \ \varphi_y (0,2) = - \frac{1}{1} = -1

[thorny flame]

Renato

[burnt swift]

$\pdv{h}{x} = \pdv{\varphi}{s} \cdot 1 + \pdv{\varphi}{t} \cdot 1$ and $\pdv{h}{y} = \pdv{\varphi}{s} \cdot 1 + \pdv{\varphi}{t} \cdot (-1)$ \ $\pdv{h}{x} = \pdv{\varphi}{s} \cdot 1 + \pdv{\varphi}{t}$ and $\pdv{h}{y} = \pdv{\varphi}{s} -\pdv{\varphi}{t}$ \ $h_x(1,-1) = \varphi_s(0,2) \cdot 1 + \varphi_t(0,2)$ and $h_y(1,-1) = \varphi_s(0,2) -\varphi_t(0,2)$ \ $\pdv{h}{x} \ (1,-1) = 6 -1$ and $\pdv{h}{y} \ (1,-1) = 6 +1$

[prisma mesa]

Seems like it's gonna be ||(5, 7)||

[burnt swift]

i mean, all of this at (0,2,-2)

[prisma mesa]

not really

it's at (s, t) = (0, 2)

but at (x, y) = (1, -1)

note that when (x, y) = (1, -1) (now talking about the b) part x, y), we have (s, t) = (x + y, x - y) = (0, 2)

[burnt swift]

h(1,-1) = phi(0,2)

[thorny flame]

Renato

[burnt swift]

is confusing exercise isnt it

[prisma mesa]

yeah, pretty confusing

[burnt swift]

all this x,y and s,t bullshit

[prisma mesa]

it's just a lot of concepts combined into one exercise, making it quite confusing

[burnt swift]

jajaj

my exam is tomorrow

just hope I will be able to pass

[prisma mesa]

is this a past exam question?

[burnt swift]

and that this doesnt happen

[burnt swift]

yeah

[prisma mesa]

yeah, good thing is that you'd probably notice it during the substitution phase

that error

but generally, the longer the sol is, the higher is the probabibility of making some arithmetic blunder

[burnt swift]

im not too worried about partial derivatives computations

but more of this phi(s,t) = phi(x,y) type of thing

[prisma mesa]

in general, it's just that
derivative of function with function inputs is the derivative of inputs * derivative of the function with normal inputs

and ofc u do that for all the partial derivatives

that's just the multivar chain rule

it surely can be confusing though

[burnt swift]

there is another way aswell

[prisma mesa]

sometimes it's better to stop looking at the individual letters and look at the bigger picture like this

[burnt swift]

,, \nabla h(1,-1) = \nabla \varphi (0,2) \cdot \left[\varphi(x,y) \right]^{'}_{(x,y) = (0,2)}

[prisma mesa]

what does [phi(x, y)] mean

[thorny flame]

Renato

[prisma mesa]

hmm

[burnt swift]

i might be tripping

[prisma mesa]

this seems to completely ignore the x + y and x - y

and what's that derivative wrt to?

[burnt swift]

you are right

anyways there is most likely a way to express it with gradients

but im not too sure tbh

so $\pdv{h}{x} = \pdv{\varphi}{s} \cdot \pdv{s}{x} + \pdv{\varphi}{t} \cdot \pdv{t}{x}$ and $\pdv{h}{y} = \pdv{\varphi}{s} \cdot \pdv{s}{y} + \pdv{\varphi}{t} \cdot \pdv{t}{y}$

[thorny flame]

Renato

[burnt swift]

tô express this i mean

[prisma mesa]

dphi / ds * gradient s + dphi/dt * gradient t

[burnt swift]

(hx, hy) = (phi_s, phi_t) * (s_x, t_x)

[prisma mesa]

this cant be right

dot product gives a scalar

while the lhs is a vector

[burnt swift]

grad h(x,y) = (phi_s, phi_t) * (s_x, t_x)

something among those lines

[burnt swift]

maybe I am tripping jajaj

[prisma mesa]

this misses s_y and t_y

[burnt swift]

my bad

[prisma mesa]

it's probably best to use this version

[burnt swift]

yeah

I'm just saying the chain rule generally speaking can be represented with gradient and dot product

[prisma mesa]

this one is a bit worse, because the normal multivar chain rule differentiates a function of one variable t

h(x(t), y(t)) and the MCR finds dh/dt (this time with normal d)

here, we have multivar function with multivar function inputs

[burnt swift]

or maybe not gradient. but jacobians

[prisma mesa]

that could be possible

[burnt swift]

yeah mb

the gradient is a special case of the Jacobian

[prisma mesa]

(s(x, y), t(x, y)) should have a 2 x 2 jacobian

[burnt swift]

you mean the jacobian for phi?

[prisma mesa]

now if we dot that with the grad phi, it could in theory work

[prisma mesa]

no, phi has just a basic gradient

[burnt swift]

math is crazy

[prisma mesa]

phi is R^2 -> R

[burnt swift]

I see, I will see this as food for thought because, like, chain rule is also like quite messy

[prisma mesa]

[
J(x,y) =
\begin{bmatrix}
\frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \
\frac{\partial t}{\partial x} & \frac{\partial t}{\partial y}
\end{bmatrix}
]

[thorny flame]

MathIsAlwaysRight

[prisma mesa]

and if we multiply this by (dphi / ds, dphi / dt), i.e. the gradient of phi

okay nvm it doesnt work

maybe transposing the jacobian would work

but its getting messy already

[burnt swift]

lmfao

[prisma mesa]

this seems like the least messy version

[burnt swift]

all this chain rule thingies are messy asfk

specially if you include implicit function theorem

[prisma mesa]

yeah, J^T * grad phi works

[burnt swift]

but yeah, life is shitty

[prisma mesa]

chain rule is quite messy

[burnt swift]

ok

maybe is just Jac(phi) which is the transpose of the gradient of phi, times J which J is the jacobian of (x+y, x-y)

[prisma mesa]

Jac(phi) doesnt make sense

phi is R^2 -> R

it only has gradient

you'd need sth R^2 -> R^2 to have a 2x2 jacobian

[burnt swift]

I mean jac(phi) is just a 2x1 jacobian

but yeah, as I said, gradient is a special case of the jacobian

[prisma mesa]

well, that makes sense

[prisma mesa]

hmm

well, that would work

Jac(phi) * Jac(s(x, y), t(x, y)) works

interesting

[burnt swift]

https://www.math.utoronto.ca/courses/mat237y1/20199/notes/Chapter2/S2.3.html

[burnt swift]

is because chain rule can be represented as multiplication of jacobians ig

[prisma mesa]

Yeah, it certainly can, i just completely forgor how its done

[burnt swift]

but yeah, multivariable calc is just a small subset of rn->rm generalisations

[prisma mesa]

now i know ig

[burnt swift]

I appreciate the convo + the help, meth

[prisma mesa]

np

[burnt swift]

just, hope I will be able to clutch my exam, I appreciate it 🙏

.solved

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Channel closed

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Use .reopen if this was a mistake.

[prisma mesa]

Good luck

[burnt swift]

ty

[topaz sinew]

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[fathom galleon]

Is my number 7 right

[finite storm]

with positive exponents

also

3x^2 - 4x - 15 can be factored

[fathom galleon]

nothing multiplies to -45 and adds to -4

😭

or am i going insane

[mild saddle]

-15*

[fathom galleon]

huh

[mild saddle]

are you using ac method here or what?

because i found a clean factorization for this

[fathom galleon]

yeah

[fathom galleon]

how would you do it?

[mild saddle]

(3x + 5)(x - 3)

but as for how i found this (which just dawned on me as to what you might be asking)

i know the factorization has to have a 3x and an x

because that's the only way you can split 3x^2

and 15 only has four factors: 1, 3, 5, 15

[fathom galleon]

so guess and check basically?

[mild saddle]

not really

well, yes but no in a way

it's an educated guess

i know that it can't be 15 and 1 because the difference is far too big

so it must be 5 and 3

and one of them has to be negative

since our middle term is negative, we know the negative one has to be bigger

[fathom galleon]

Alright yeah I see what your saying

[mild saddle]

then i kind of noticed that if i take 3 to pair it with the 3x, 3x3 = 9, and the other term is 5, and we can make 4 with 9 and 5

...number sense, I suppose?

but if ever in doubt

just throw the quad formula at it

[fathom galleon]

Does number 6 look good to you my final question

[fathom galleon]

👍

[mild saddle]

looks right to my tired eyes

[fathom galleon]

my tired eyes say it too but the program is marking it wrong

ive checked this one 10000 times

idk

[mild saddle]

let me have a closer look then

i have no idea what the page wants then

[fathom galleon]

thaanks for trying i'll js talk to my teacher abt it

[mild saddle]

aight

mb for not being able to find out what the page wants

[fathom galleon]

nws lol

ty for helping me

.close

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[autumn plaza]

how do ido this?

[autumn plaza]

I’ve done this so far

[topaz sinew]

@autumn plaza Has your question been resolved?

[topaz sinew]

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[sick hawk]

hey guys super easy question here but mark scheme says im wrong

[sick hawk]

a= 28

which is correct

so i set the equation

0 = 1/2 * k * (56+(k-1)(-4))

is this correct

oh nevermind sorry guys

.close

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[hallow rain]

does anyone know how to do something like ths

[sour bloom]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sour bloom]

do you know Bellman-Ford or Ford-Fulkerson

or Dinic or push-relabel?

[hallow rain]

ive been trying to research the ford-fulkerson method

ive watched 2 videos

one of them makes a lot of sense and one does not

they seem to have different methods

I have not begun

[sour bloom]

So one way to find a flow is repeat:

1. find path from A to B
2. send flow along path
3. update graph for residual flows

[hallow rain]

yeah so the edge with the lowest flow would be like the limiting factor right?

so for example A-C-D-B would send 11 units of flow

https://cdn.discordapp.com/attachments/903486479064522752/1421110493556379698/image.png?ex=68d7d7c8&is=68d68648&hm=b99abb4b0b3261d20672923ac701cd93020893e0d6232551c69df7ba32b4534a&

[sour bloom]

yeah you can send 11 units of flow since you see it, then you update the graph

[hallow rain]

and then a-e-c-d-b would send 2 units of flow?

and then a-e-f-b would send 7, and a-e-c-d-f-b would send 1?

and the total would be 20? thats my ucrrent conclusion

would the maimum data throughput just be like hte max that would make it to B?

[sour bloom]

you can check with maxflow-mincut theorem if you aren't sure

[hallow rain]

what do you do know as a postgrad in maths?

now*

[sour bloom]

like lots of undergrad stuff

[hallow rain]

lemme try this and see where i end up real quick

[sour bloom]

i'm not seeing any min cut of 20, did you update your flows correctly

this should be your updated graph after sending 11 units of flow

[hallow rain]

dont they only have 1 direction to them?

[sour bloom]

after you send 11 units of flow forwards, you may later regret sending forwards and may choose to send back instead

that's how you update your graph

[hallow rain]

i thought like the arrows mean that it can only flow one way?

[sour bloom]

yeah i think you just made a calculation error

[sour bloom]

yeah, you can send between 0 and maximum capacity

but after you send 11, what you are allowed to do is send up to 11 backwards or send up to "maximum capacity-11" forwards

[hallow rain]

quick question: would the absolute max that can go to B be 21? because the 2 paths leading to b add up to 21?

[sour bloom]

what you saw was the edges 14 and 7 form a cut of the graph, yeah?

and 14+7=21

13 + 9 = 22 btw

[hallow rain]

oh

mb

so then my original answer shouldve been 21

but let me finish drawing out what im doing rq

sorry so would the absolute max be 22?

[sour bloom]

well, the ingoing edges into B would form an upper bound, but it's not a tight upper bound

[hallow rain]

Sorry if its a bit messy but that’s basically what the first video i watched had done

[sour bloom]

yeah you can notate like that too

[hallow rain]

does that look remotely right? my work is due in 1 hour and 11 minutes and i cant be bothered anymore with this question 😔😔

as long as i get some marks

just gotta annotate it

[sour bloom]

may want to indicate next to each diagram how much flow you sent, to be slightly clearer

[hallow rain]

yeah

i gotta explain it still

[sour bloom]

• you can try to explain why you cannot send any more flow

[hallow rain]

ok ill make sure to add that

[sour bloom]

those are the quickest improvements to make

[hallow rain]

beautiful thank you

i guess i am done

thank you for your help

🙏

.close

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[willow nest]

helpp

[willow nest]

Two important players in a football team are injured, and each has probability 1/3 ofrecovering before the next match. The recoveries of the two players are independent of eachother. If both are able to play then the team has probability 3/4 of winning the match,if only one of them plays then the probability of winning is 1/2 and if neither play theprobability of winning is 1/16. What is the probability that the team wins the match?

[true magnet]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[willow nest]

idk what to do

[true magnet]

what are the possible outcomes

[true magnet]

consider the possible outcomes of P1 and P2 recovering

[willow nest]

if p1 recoveres then
1/2 chance winning
if p1 and p2 recovered then 3/4 winnning

if neither recover then 1/16 winning

[true magnet]

what's the probability of each of those

[true magnet]

ok, what's the probability of neither recovering

[willow nest]

8/9

wait i got it

.close

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[merry helm]

how do i do this fast

[merry helm]

scores are bigger on x axis so i assume x axis is written and y axis is practical

lowest score on practical is Ina

then slighly higher score, but less on the written is Liz

idk what to do now, do i just verify each point 1 by 1

[balmy plank]

I would check 1 axis and if something seems off cross-check with the other axis.

[merry helm]

nvm im dumb, i can only check the ones which there are options for

.close

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[robust plover]

Similar to #help-8 I just want to know if the solution here is motivated/well explained.

[topaz sinew]

Please don't occupy multiple help channels.

[clear python]

!1c please

[topaz sinew]

Please stick to your channel.

[robust plover]

I should send multiple question to the same channel ?

ohh

Ok lemme close then

[clear python]

Wait for one question to be resolved

Then send the next

[robust plover]

.close

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[robust plover]

Ok thanks

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[devout veldt]

How can i find the value of this

[shut obsidian]

Well, write out what this means

[devout veldt]

1/19(sinx/cos19)?

[shut obsidian]

I didn't mean that

|_a^b is evaluation at b - evaluation at a usually

I assume that's what it is in your context also?

[devout veldt]

i genuinely dont understand what you meaN

[raven field]

First off do you know
lim x->inf tan^-1(x) ?

[devout veldt]

no

[shut obsidian]

$f(x) \Big|_a^b = f(b) - f(a)$, right?

[thorny flame]

Kepe

[devout veldt]

right

[shut obsidian]

So $\frac{1}{12} \tan^{-1} \l(\frac{x}{12}\r) \Big|_a^0 = \frac{1}{12} \tan^{-1} \l(0\r) - \frac{1}{12} \tan^{-1} \l(\frac{a}{12}\r)$, right?

[thorny flame]

Kepe

[devout veldt]

im following

[shut obsidian]

And tan^(-1)(0) = 0

[devout veldt]

correct

[shut obsidian]

So $\frac{1}{12} \tan^{-1} \l(0\r) - \frac{1}{12} \tan^{-1} \l(\frac{a}{12}\r) = - \frac{1}{12} \tan^{-1} \l(\frac{a}{12}\r)$

[thorny flame]

Kepe

[shut obsidian]

And in total that gives us [\lim_{a \to -\infty} \l(\frac{1}{12} \tan^{-1} \l(\frac{x}{12}\r)\r) \Big|a^0 = \lim{a \to -\infty} - \frac{1}{12} \tan^{-1} \l(\frac{a}{12}\r) = - \frac{1}{12} \lim_{a \to -\infty} \tan^{-1} \l(\frac{a}{12}\r)]

[devout veldt]

so how do I find the value of that

[thorny flame]

Kepe

[shut obsidian]

As a -> -infinity, what does a/12 approach

[devout veldt]

infinity

[shut obsidian]

Yeah almost

-infinity

[devout veldt]

right

[shut obsidian]

I.e. what we have is just [\lim_{a \to -\infty} \l(\frac{1}{12} \tan^{-1} \l(\frac{x}{12}\r)\r) \Big|a^0 = \lim{a \to -\infty} - \frac{1}{12} \tan^{-1} \l(\frac{a}{12}\r) = - \frac{1}{12} \lim_{a \to -\infty} \tan^{-1} \l(\frac{a}{12}\r) ][= - \frac{1}{12} \lim_{a \to -\infty} \tan^{-1} \l(a\r) = - \frac{1}{12} \cdot \l(-\frac{\pi}{2}\r)]

Now we need to find out what arctan approaches as it goes to -infinity

[devout veldt]

yeah

[shut obsidian]

Do you know how the graph of tan looks like?

,w plot tan(x)

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kepe_5235

[shut obsidian]

For arctan we only consider tan(x) in the interval -pi/2 < x < pi/2

[devout veldt]

alright

[shut obsidian]

,w plot tan(x), -pi/2 < x < pi/2

[shut obsidian]

i.e. for arctan we only consider the thing between the red

Now tan^(-1) is the inverse function of tan

I.e. it's the x value for the y that y = tan(x) would give us

[shut obsidian]

We want to think about what value tan^(-1) approaches as x-> -infininity, i.e. what x we need to approach with tan for tan(x) to go to -infinity

You can read that off the graph now

(we are only in the interval (-pi/2, pi/2))

[devout veldt]

-pi/2

right

[shut obsidian]

Right

[thorny flame]

Kepe

[devout veldt]

so pi/24

ahhhhh

[shut obsidian]

Yeah

[devout veldt]

thank you for going in depth

this was all very helpful

[shut obsidian]

np

[devout veldt]

that's basically everything right

[shut obsidian]

Yeah

gfauxpas might have something to still say

[neat jasper]

arctan is a very useful function because it is a one to one correspondence between (-infinity,+infinity) and (-pi/2,pi/2)

keep its domain and range in mind for the future

for example

do you know what a distance function is?

[devout veldt]

distance function?

[neat jasper]

yeah

also called a metric

[devout veldt]

I dont think so

[neat jasper]

it's a function that takes two objects and returns a notion of "distance" between them

the most commonly used one is the one from the pythagorean theorem in the plane

d((x1,y1),(x2,y2)) = sqrt(delta x^2 + delta y^2)

stolen from google image search

okay nevermind, if you havent seen this before it's too complicated

[shut obsidian]

Well what was probably the intention is that one can show that tan^(-1) is actually a metric!

But yeah probably not worth the effort now if you haven't had real analysis yet lol

[neat jasper]

okay let me give a better example that doesnt require analysis

let's say you want to think about the interval (-infinity,+infinity) but

you want to think about what would happen if you treated infinity as numbers, what would change? does it make sense to have the interval [-infinity, +infinity]?

well, that's a complicated question

but

a much easier question is

what happens if you take the interval (-pi/2, pi/2) and add two points to the ends so you get [-pi/2, pi/2]?

that's a lot more tractable

[devout veldt]

yeah

[neat jasper]

so let's say I want to ask you

if f(x) = 1/x, what is f(+infinity)?

you intuitively want it to be zero

so you can do things like

let f(theta)=1/tan(theta) as theta in [-pi/2, pi/2] and it's equivalent to examine what happens in that case

i picked an example that actually wasn't a good example lol

but the idea is correct

did you do integrals yet?

[shut obsidian]

Concretely, arctan gives you a one-to-one 'identification' (bijection) of R with the interval (-pi/2, pi/2) which might sound surprising/is nice because one of them is infinite and the other one is finite

[neat jasper]

if you know integration by substitution I can give you a classic example

[devout veldt]

I'm doing integral problems right now

and this just happens to be the last one on my homework

[neat jasper]

okay let me show you a classic then

$$\int_{-\infty}^{+\infty} \frac{\mathrm dx}{1+x^2}$$

[thorny flame]

gfauxpas

[neat jasper]

not immediately obvious how to do this, probably

but

let's substitute something that turns this integral from one over (-infinity,+infinity) to one instead over (-pi/2,pi/2)

Let's let integration by substitution and say

x = tan θ, dx = sec^2 θ dθ

do you see how the new bounds of integration become -pi/2 to pi/2?

[devout veldt]

so basically a shortcut

[neat jasper]

eh, not really a shortcut, it's more of

a trick

a nice trick

$$\int_{-\pi/2}^{\pi/2} \dfrac{\sec^2 \theta , \mathrm d\theta}{1+\tan^2 \theta}$$

[thorny flame]

gfauxpas

[neat jasper]

what next, any suggestions?

[devout veldt]

not really

all pretty helpful

[neat jasper]

do you remember that

$$\sin^2 \theta + \cos^2 \theta = 1$$

[thorny flame]

gfauxpas

[devout veldt]

yeah

[neat jasper]

what happens if you divide all terms by cos^2 theta

[devout veldt]

tan^2theta + 1 = sec^2theta

[neat jasper]

tan^2 θ + 1 = sec^2 θ you mean

[devout veldt]

yeah

[neat jasper]

ah, look at that! th enumerator of the integrandd is the same as the denominator!

so it's just

$$\int_{-\pi/2}^{\pi/2} , \mathrm d \theta$$

[thorny flame]

gfauxpas

[devout veldt]

so just theta then

[neat jasper]

and we said

x = tan theta

[devout veldt]

which is pi

right

[neat jasper]

so theta = arctan x

so we have

well

i think you can finish it :)

but neat, right?

[devout veldt]

pretty cool yeah

i'll let the channel be free

.solved

[topaz sinew]

Channel closed

Closed by @devout veldt

Use .reopen if this was a mistake.

[topaz sinew]

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• Show your work, and if possible, explain where you are stuck.
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[weak plank]

Question on an past admission exam im trying to get on it translates to:

If X = 3 + 3 square root
Whats X³ then

[ruby tree]

More than a translation, we'd like to know what you tried

[weak plank]

sure

ill break down what i did

i solved the equation yesterday with a little help from my friend BUT I FORGOT HOW TO and the result is 54 + 30 3 square root

i realized just X³ straight up the whole thing which results in 27 + 3 3 square root

which i realized is not quite right

[weak plank]

aditionally

[ruby tree]

Yeah, that's right

[odd pagoda]

just like (a+b)^2 isnt a^2+b^2, (a+b)^3 also isnt a^3+b^3

[weak plank]

i used the "breakdown method"

help me translate im stupid i dont know english formula names

[weak plank]

this

but i tried doing this again but i got different results

[weak plank]

correct thats why i realizes its not quite right

[ruby tree]

The formula for (a+b)^2 is called an (algebraic) identity

[weak plank]

ill show my though process summed up and you guys tell me what i did wrong

[weak plank]

ok ok right thanks i used this

[ruby tree]

(a+b)^3 also has an identity, but less commonly known

[weak plank]

So like if i wanna get it right i should do (A + B)² which is

A² + 2AB + B²

[ruby tree]

Right

Though the question is x^3 not x^2

[weak plank]

then this times this A + B

[weak plank]

right

[ruby tree]

Yes

[weak plank]

sure this clears things up

should be pretty straightfoward ill tell you the results in a secound

ok this really clears things up guys thanks

same result again but i know how to do it now

54 + 30 square root of 3

.close

[topaz sinew]

Channel closed

Closed by @weak plank

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[topaz sinew]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[heavy timber]

Why is the angle between two planes (in 3D) = the angle between their perpendicular vectors?

[heavy timber]

I cant really imagine it? when i try to draw it, i get a rhombus and the angles are added up to 180 but not equal.

But i drew it in a 2d way so I might be very wrong.

[exotic dome]

It’s the angle between their normal vectors

[loud oasis]

same thing?

[shut obsidian]

Visualize two planes intersecting and visualize their normal vectors. Now at the same time imagine their normal vectors tilting 90° into the plane

[heavy timber]

wdym "normal vectors"

heres what i drew

yellow and white are planes

x is the angle between them

[exotic dome]

Yes, thats what I said

[heavy timber]

how are the x's even equal.

[exotic dome]

It’s the easiest way to represent it

[shut obsidian]

I don't think it's very good to visualize this in 2d though, let me see if I can find a 3d one

[heavy timber]

please do!

[heavy timber]

as i cannot imagine what you're trying to say here

[loud oasis]

even in 3D you can always look at it "side-on" to get the same view you are getting here

[heavy timber]

isnt "alpha + beta" = 180 just a quadrilateral shape property

two opposite angles add up to 180